链接:Link 状态:Runtime Error

前言

这题做的可真够久的,整整三个小时。但即便如此,还是只过了一部分的点,另一部分报运行时错误——估计是哈希表设计的不太好。但这确实是一道好题,因此,在睡觉前决定记录一下。

分析

很容易便想到:用一个三元组 $(x,y,z)$ 表示节点,表示内容为 x 的节点下跟着标号为 y 和 z 的左右子树。这样一来,一类相同的子树便可以唯一确定了,而不必每构造一棵子树就把整棵树遍历一遍。
对于三元组的储存,刚开始图方便,用了数组。查找也是用了 $O(n)$ 的线性查找。磕磕碰碰写了两个多小时然后兴冲冲地提交,结果TLE了…………没办法,只好又花了半个小时写了一个哈希表,然后就是上文说过的情况了:Runtime Error204。可能是哈希数组过大的原因,日后再微调一下,今天实在是没有脑子了。

Code

const
maxn = 20000;
type
NodeRec = record
Value: string;
l, r, index: longint;
end;
Node = record
left, right: longint; //Index of left and right child in the `tree` array, -1 for none.
Rec: NodeRec;
index: longint;
end;
//以下为哈希表的定义
_PNode = ^_Node;
_Node = record
n: Node;
next: _PNode;
end;

HashTable = object
arr: array [0..maxn] of _PNode;
function hash(n: NodeRec): longint;
procedure add(n: Node);
procedure clear;
function find(n: NodeRec): longint;
end;

procedure HashTable.clear;
var
i: longint;
p, q: _PNode;
begin
for i := 0 to maxn do
begin
p := arr[i];
while p<>nil do
begin
q := p^.next;
dispose(p);
p := q;
end;
end;
fillchar(arr, sizeof(arr),0);
end;

function cmp(r1, r2: NodeRec): Boolean;
begin
cmp := (r1.l = r2.l) and (r1.r = r2.r) and (r1.Value = r2.Value);
end;

function HashTable.hash(n: NodeRec): longint;
var
i: longint;
begin
hash := 0;
for i := 1 to length(n.Value) do
hash := (hash*5 + ord(n.Value[i]) - ord('a')) mod maxn;
hash := (hash + n.l * 10 + n.r * 5) mod maxn;
end;

procedure HashTable.add(n: Node);
var
h: longint;
p, q: _PNode;
begin
h := hash(n.rec);
new(q);
fillchar(q^, sizeof(_Node), 0);
q^.next := arr[h];
q^.n := n;
arr[h] := q;
end;

function HashTable.find(n: NodeRec): longint;
var
p: _PNode;
begin
find := -1;
p := arr[hash(n)];
while (p<>nil) and not cmp(n, p^.n.rec) do p := p^.next;
if p <> nil then
find := p^.n.index;
end;
//哈系表定义结束
var
inputs: Ansistring;
_: longint;
tree: array [1..50001] of Node;
cur: longint; //The current pointer of the input string.
num: longint; //The current number of the `tree` array.
ls: longint;
t: longint;
tot: longint;
ht: HashTable;

function build: longint; //建树
label lb;
var
rec: NodeRec;
i,j,l,r: longint;
begin
l := 0;
r := 0;
fillchar(rec, sizeof(rec), 0);
inc(tot);
rec.index := tot;
while (cur<=ls) and (inputs[cur] in ['a'..'z']) do
begin
rec.Value := rec.Value+inputs[cur];
inc(cur);
end;
if cur>ls then goto lb; //。。。这里被迫跳转控制流,由于实在不想多谢,就用了臭名昭著的label
if inputs[cur] = '(' then
begin
inc(cur);
l := build();
rec.l := tree[l].rec.index;
inc(cur);
r := build();
rec.r := tree[r].rec.index;
inc(cur);
end;
j := ht.find(rec);
if j>0 then
begin
dec(tot);
exit(j);
end
else
begin
lb:
inc(num);
tree[num].left := l;
tree[num].right := r;
tree[num].rec := rec;
tree[num].index := num;
ht.add(tree[num]);
exit(num);
end;
end;

procedure print(n: longint);
begin
if tree[n].rec.index > t then
begin
write(tree[n].rec.Value);
t := tree[n].rec.index;
end
else
begin
write(tree[n].rec.index);
exit;
end;
if tree[n].right = 0 then
exit;
write('(');
print(tree[n].left);
write(',');
print(tree[n].right);
write(')');
end;
begin
assign(input, 'main.in'); reset(input);
assign(output, 'main.out'); rewrite(output);
readln(_);
fillchar(ht.arr, sizeof(ht.arr),0);
while _>0 do
begin
dec(_);
readln(inputs);
fillchar(tree, sizeof(tree), 0);
ht.clear;
ls := length(inputs);
cur := 1; num := 0; tot := 0;
build;
t := 0;
print(num);
writeln;
end;
close(input); close(output);
end.